When Right Angle Is Not

Given,

 $x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}$

what happens when $v\ge1$?

When $x^{'}$ was set to $1$ in the triangle (post "Thought Experiment Gone Wrong" dated 20 Nov 2017),

$cos(\theta)=\cfrac{\sqrt{1-v^2}}{1}$

we are attempting to transform $x^{'}$ to $x$ where it is calibrated to unit length.  We are saying if $x^{'}$ is to be one unit what do we need to do to $x$.  Using the triangle restricts the value of $v$ to less than $1$.

So, given $v$ and $x$ and $v\ge 1$ to use the transform, both $v$ and $x$ are scale up to the value of the highest speed possible, light speed $c$ and $|c|$ respectively;

$\cfrac{v}{c}\rightarrow v$

$\cfrac{x}{|c|}\rightarrow x$

and $x^{'}$ is also measured in the unit of $c$ per unit time.

$\cfrac{x^{'}}{|c|}\rightarrow x^{'}$

And hence Lorentz's transform.

Note: If $x=10\,m$ then to used Lorentz transform we need $\cfrac{10}{299792458}\rightarrow x$ instead of just $x=10$.